1. Introduction A Closer Look at Graphing The Distance Formula
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Introduction
A Closer Look at Graphing
Translations
Definition of a Circle
Definition of Radius
Definition of a Unit Circle
Deriving the Equation of a Circle
The Distance Formula
Moving The Center
Completing the Square
It is a Circle if?
Conic Movie
Conic Collage
Geometers Sketchpad: Cosmos
Geometers Sketchpad: Headlights
Projectile Animation
Planet Animation
Plane Intersecting a Cone Animation
Footnotes
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2. CONIC SECTIONS Quadratic Relations Parabola Circle Ellipse Hyperbola
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3. The quadratic relations that we studied in the beginning
of the year were in the form of y = Ax2 + Dx + F,
where A, D, and F stand for constants, and A ≠ 0.
This quadratic relation is a parabola, and it is the only one that can be a function.
It does not have to be a function, though.
A parabola is determined by a plane intersecting a cone and is therefore considered a conic section.
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4. The general equation for all conic sections is:
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
where A, B, C, D, E and F represent constants
and where the equal sign could be replaced by an
inequality sign.
When an equation has a “y2” term and/or an “xy” term it is a quadratic relation instead of a quadratic function.
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5. A Closer Look at Graphing Conics
Plot the graph of each relation. Select values of x and
calculate the corresponding values of y until there are
enough points to draw a smooth curve. Approximate all
radicals to the nearest tenth.
x2 + y2 = 25
x2 + y2 + 6x = 16
x2 + y2 - 4y = 21
x2 + y2 + 6x - 4y = 12
Conclusions
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6. x2 + y2 = 25
x = 3
32 + y2 = 25
9 + y2 = 25
y2 = 16
y = ± 4
There are 2 points to graph:
(3,4)
(3,-4)
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A Close Look at Graphing
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7. Continue to solve in this manner, generating a table of values.-5 -4 ±3 -3 ±4 -2
±4.6 -1
±4.9 ±5 1
±4.9 2
±4.6 3 ±4 4 ±3 5
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8. Graphing x2 + y2 = 25 y x End A Close Look at Graphing Main Menu
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9. x2 + y2 + 6x = 16 x = 1 12 + y2 +6(1) = 16 1 + y2 +6 = 16 y2 = 9
There are 2 points to graph:
(1,3)
(1,-3)
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A Close Look at Graphing
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10. Continue to solve in this manner, generating a table of values.-8 -7 ±3 -6 ±4 -5
±4.6 -4
±4.9 -3 ±5 -2 -1 1 2 -8 -7 ±3 -6 ±4 -5
±4.6 -4
±4.9 -3 ±5 -2
±4.9 -1
±4.6 ±4 1 ±3 2
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11. Graphing x2 + y2 + 6x = 16 -8 -7 ±3 -6 ±4 -5 ±4.6 -4 ±4.9 -3 ±5 -2 -1-7 ±3 -6 ±4 -5
±4.6 -4
±4.9 -3 ±5 -2 -1 1 2
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12. x2 + y2 - 4y = 21
x = 3
32 + y2 - 4y = 21
9 + y2 - 4y = 21
y2 - 4y -12 = 0
(y - 6)(y + 2) = 0
y - 6 = 0 and y + 2 = 0
y = 6 and y = -2
There are 2 points to graph:
(3,6)
(3,-2)
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13. Continue to solve in this manner, generating a table of values.
-3, 7 3
-2, 6 4
-1, 5 5 2 1
-2.9, 6.9
-2.6, 6.6 -5 -4 -3 -2 -1 -5 2 -4
-1, 5 -3
-2, 6 -2
-2.6, 6.6 -1
-2.9, 6.9
-3, 7 1
-2.9, 6.9 2
-2.6, 6.6 3
-2, 6 4
-1, 5 5 2
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A Close Look at Graphing
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14. Graphing x2 + y2 - 4y = 21
-3, 7 3
-2, 6 4
-1, 5 5 2 1
-2.9, 6.9
-2.6, 6.6 -5 -4 -3 -2 -1
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15. x2 + y2 + 6x - 4y = 12
x = 1
12 + y2 + 6(1) - 4y = 12
1 + y y = 12
y2 - 4y - 5 = 0
(y - 5)(y + 1) = 0
y - 5 = 0 and y + 1 = 0
y = 5 and y = -1
There are 2 points to graph:
(1,5)
(1,-1)
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A Close Look at Graphing
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16. Continue to solve in this manner, generating a table of values.-8 2 -7
-1, 5 -6
-2, 6 -5
-2.6, 6.6 -3
-3, 7 -2
-2.9, 6.9 -1 -4 1 -8 2 -7
-1, 5 -6
-2, 6 -5
-2.6, 6.6 -4
-2.9, 6.9 -3
-3, 7 -2
-2.9, 6.9 -1
-2.6, 6.6
-2, 6 1
-1, 5 2
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17. Graphing x2 + y2 + 6x - 4y = 21 -8 2 -7 -1, 5 -6 -2, 6 -5 -2.6, 6.6 -3
-3, 7 -2
-2.9, 6.9 -1 -4 1
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18. What conclusions can you draw about the shape and location of the graphs?
x2 + y2 = 25
x2 + y2 + 4x = 16
x2 + y2 + 4x - 6y = 12
x2 + y 2 - 6y = 21
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19. What conclusions can you draw about the shape and location of the graphs?
All of the graphs are circles x2 + y2 = 25
As you add an x-term, the graph moves left x2 + y2 + 4x = 16
As you subtract a y-term, the graph moves up. x2 + y 2 - 6y = 21
You can move both left and up x2 + y2 + 4x - 6y = 12
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20. GEOMETRICAL DEFINITION OF A CIRCLE
A circle is the set of all points in a plane equidistant from a fixed point called the center.
Center
Circle
Circle
Center
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21. A radius is the segment whose endpoints are the center of the circle, and any point on the circle.
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22. It is the circle on which all other circles are based.
A unit circle is a circle with a radius of 1 whose center is at the origin.
r = 1
A unit circle is a circle with a radius of 1 whose center is at the origin.
Center: (0, 0)
A unit circle is a circle with a radius of 1 whose center is at the origin.
A unit circle is a circle with a radius of 1 whose center is at the origin.
It is the circle on which all other circles are based.
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23. The equation of a circle is derived from its radius.
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24. Use the distance formula to find an equation for x and y
Use the distance formula to find an equation for x and y. This equation is also the equation for the circle.
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Deriving the Equation of a Circle
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25. THE DISTANCE FORMULA (x2, y2) (x1, y1) End
Deriving the Equation of a Circle
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26. Let r for radius length replace D for distance.
(x, y)
(0, 0)
r2 = x2 + y2
r2 = x2 + y2
Is the equation for a circle with its center at the origin and a radius of length r.
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Deriving the Equation of a Circle
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27. The unit circle therefore has the equation:
x2 + y2 = 1
(x, y)
(0, 0)
r = 1
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Deriving the Equation of a Circle
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28. (x, y) r = 2 If r = 2, then (0, 0) x2 + y2 = 4 End
Deriving the Equation of a Circle
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29. In order for a satellite to remain in a circular orbit above the Earth, the satellite must be 35,000 km above the Earth.
Write an equation for the orbit of the satellite. Use the center of the Earth as the origin and 6400 km for the radius of the earth.
(x - 0)2 + (y - 0)2 = ( )2
x2 + y2 = 1,713,960,000
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Deriving the Equation of a Circle
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30. What will happen to the equation if the center is not at the origin?
(1,2)
r = 3
(x,y)
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31. No matter where the circle is located, or where the center is, the equation is the same.
(h,k)
(x,y) r
(h,k)
(x,y) r
(h,k)
(x,y) r
(h,k)
(x,y) r
(h,k)
(x,y) r
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Moving the Center
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32. Assume (x, y) are the coordinates of a point on a circle.
(h,k)
The center of the circle is (h, k), r
and the radius is r.
Then the equation of a circle is: (x - h)2 + (y - k)2 = r2.
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Moving the Center
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33. Write the equation of a circle with a center at (0, 3) and a radius of 7.
(x - h)2 + (y - k)2 = r2
(x - 0)2 + (y - 3)2 = 72
(x)2 + (y - 3)2 = 49
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34. Find the equation whose diameter has endpoints of (-5, 2) and (3, 6).
First find the midpoint of the diameter using the midpoint formula.
This will be the center.
MIDPOINT
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35. Find the equation whose diameter has endpoints of (-5, 2) and (3, 6).
Then find the length distance between the midpoint and one of the endpoints.
This will be the radius.
DISTANCE FORMULA
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36. Find the equation whose diameter has endpoints of (-5, 2) and (3, 6).
Therefore the center is (-1, 4)
The radius is
(x - -1)2 + (y - 4)2 =
(x + 1)2 + (y - 4)2 = 20
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Moving the Center
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37. A line that intersects the circle in exactly one
A line in the plane of a circle can intersect the circle in 1 or 2 points.
A line that intersects the circle in exactly one
point is said to be tangent to the circle.
The line and the
circle are considered tangent to each other at this point of intersection.
Write an equation for a circle with center (-4, -3) that is tangent to the x-axis.
A diagram will help.
(-4, 0)
A radius is always perpendicular to the tangent line. 3
(x + 4)2 + (y + 3)2 = 9
(-4, -3)
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38. The standard form equation for all conic sections is:
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0
where A, B, C, D, E and F represent constants and where the equal sign could be replaced by an inequality sign.
How do you put a standard form equation into graphing form?
The transformation is accomplished through completing the square.
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39. Graph the relation x2 + y2 - 10x + 4y + 13 = 0.
2. Group the x-terms and y-terms together
x2 - 10x + y2 + 4y = -13
Graph the relation 13
1. Move the F term to the other side.
x2 + y2 - 10x + 4y = -13
3. Complete the square for the x-terms and y-terms.
x2 - 10x + y2 + 4y = -13
x2 - 10x y2 + 4y + 4 =
(x - 5)2 + (y + 2)2 = 16
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Completing the Square
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40. (x - 5)2 + (y + 2)2 = 16 (x - 5)2 + (y + 2)2 = 42 center: (5, -2)
radius = 4 4
(5, -2)
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Completing the Square
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41. What if the relation is an inequality? x2 + y2 - 10x + 4y + 13 < 0
Do the same steps to transform it to graphing form.
(x - 5)2 + (y + 2)2 < 42
This means the values are inside the circle.
The values are less than the radius.
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Completing the Square
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42. Write x2 + y2 + 6x - 2y - 54 = 0 in graphing form
Write x2 + y2 + 6x - 2y - 54 = 0 in graphing form. Then describe the transformation that can be applied to the graph of x2 + y2 = 64 to obtain the graph of the given equation.
x2 + y2 + 6x - 2y = 54
x2 + 6x + y2 - 2y = 54
(6/2) = (-2/2) = -1
(3)2 = (-1)2 = 1
x2 + 6x y2 - 2y + 1 =
(x + 3)2 + (y - 1)2 = 64
(x + 3)2 + (y - 1)2 = 82
center: (-3, 1) radius = 8
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Completing the Square
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43. Write x2 + y2 + 6x - 2y - 54 = 0 in graphing form
Write x2 + y2 + 6x - 2y - 54 = 0 in graphing form. Then describe the transformation that can be applied to the graph of x2 + y2 = 64 to obtain the graph of the given equation.
x2 + y2 = 64 is translated 3 units left and one unit up to become (x + 3)2 + (y - 1)2 = 64.
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Completing the Square
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44. The graph of a quadratic relation will be a circle if the coefficients of the x2 term and y2 term are equal (and the xy term is zero).
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